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222-9+11+12:2*14+14 = ? ( )

Toán Lớp 8: Tìm $x$ a, $3x$ . $(x-1)+x-1=0$ b, $2.(x+3)$ – $x^{2}$ – $3x=0$ c, $4x^{2}$ – $25$ – $(2x-5).(2x+7)$ $=0$ d, $x^{3}$ + $27$ + $(x

Home/ Toán Lớp 8: Tìm $x$ a, $3x$ . $(x-1)+x-1=0$ b, $2.(x+3)$ – $x^{2}$ – $3x=0$ c, $4x^{2}$ – $25$ – $(2x-5).(2x+7)$ $=0$ d, $x^{3}$ + $27$ + $(x

Toán Lớp 8: Tìm $x$ a, $3x$ . $(x-1)+x-1=0$ b, $2.(x+3)$ – $x^{2}$ – $3x=0$ c, $4x^{2}$ – $25$ – $(2x-5).(2x+7)$ $=0$ d, $x^{3}$ + $27$ + $(x

Tháng Mười Một 9, 2022 Toán học 2 Comments 0 views

Toán Lớp 8: Tìm $x$
a, $3x$ . $(x-1)+x-1=0$
b, $2.(x+3)$ – $x^{2}$ – $3x=0$
c, $4x^{2}$ – $25$ – $(2x-5).(2x+7)$ $=0$
d, $x^{3}$ + $27$ + $(x+3).(x-9)$ $=0$

Comments ( 2 )

  1. lanhuongthu
    lanhuongthu
    9 Tháng Mười Một, 2022 at 10:32 sáng
    Reply
    a, 3x. (x – 1) + x – 1 = 0
       3x. (x – 1) + (x – 1) = 0
       (x – 1) . (3x + 1) = 0
    $\Rightarrow$ $\left[\begin{matrix} x-1=0\\ 3x+1=0\end{matrix}\right.$
    $\Rightarrow$ $\left[\begin{matrix} x=1\\ 3x=-1\end{matrix}\right.$
    $\Rightarrow$ $\left[\begin{matrix} x=1\\ x=-1/3\end{matrix}\right.$
    Vậy x $\in$ { 1; $\frac{-1}{3}$ }
    b, 2. (x + 3) – x² – 3x = 0
        2. (x + 3) – (x² + 3x) = 0
        2. (x + 3) – x. ( x + 3) = 0
         ( x + 3) . (2 – x) = 0
    $\Rightarrow$ $\left[\begin{matrix} x+3=0\\2-x=0\end{matrix}\right.$
    $\Rightarrow$ $\left[\begin{matrix} x=-3\\ x=2\end{matrix}\right.$
    Vậy x $\in$ { -3; 2}
    c, 4x² – 25 – (2x – 5). (2x + 7) = 0
       (2x)² – 5² – (2x – 5). (2x + 7) = 0
       ( 2x – 5) . (2x + 5) – (2x – 5). (2x + 7) = 0
        (2x – 5) . [(2x + 5) – (2x + 7)] = 0
        (2x – 5) . (2x + 5 – 2x -7) = 0
        ( 2x – 5) . (-2) = 0
    $\Rightarrow$ 2x – 5 = 0
    $\Rightarrow$ 2x = 5
    $\Rightarrow$ x = 5/2
    Vậy x = $\frac{5}{2}$ 
    d, x³ + 27 + (x + 3). (x – 9) = 0
       x³ + 3³ + ( x + 3) . ( x – 9) = 0
       (x + 3). (x² – x.3 + 3²) + ( x + 3) . ( x – 9) = 0
       ( x + 3) . ( x² – 3x + 9) + ( x + 3) . ( x – 9) = 0
       ( x + 3) . [( x² – 3x + 9) + ( x – 9)] = 0
        ( x + 3) . ( x² – 3x + 9 + x – 9) = 0
        (x + 3) . ( x² – 2x) = 0
        ( x + 3). x. ( x – 2) = 0
    $\Rightarrow$ $\left[\begin{matrix} x+3=0\\ x-2=0\end{matrix}\right.$
    $\Rightarrow$ $\left[\begin{matrix} x=-3\\ x=2\end{matrix}\right.$
    Vậy x $\in$ { -3; 2 }

  2. bichhhathu
    bichhhathu
    9 Tháng Mười Một, 2022 at 10:32 sáng
    Reply
    Giải đáp + Lời giải và giải thích chi tiết:
    a, 3x(x-1)+x-1=0
    ⇔3x(x-1)+(x-1)=0
    ⇔(x-1)(3x+1)=0
    ⇔[(x=1),(x=-1/3):}
    Vậy S={1;-1/3}
    b, 2(x+3)-x^2-3x=0
    ⇔2(x+3)-x(x+3)=0
    ⇔(x+3)(2-x)=0
    ⇔[(x=-3),(x=2):}
    Vậy S={-3;2}
    c, 4x^2-25-(2x-5)(2x+7)=0
    ⇔(2x-5)(2x+5)-(2x-5)(2x+7)=0
    ⇔(2x-5)(2x+5-2x-7)=0
    ⇔(2x-5)(-2)=0
    ⇔2x-5=0
    ⇔x=5/2
    Vậy S={5/2}
    d, x^3+27+(x+3)(x-9)=0
    ⇔(x+3)(x^2+3x+9)+(x+3)(x-9)=0
    ⇔(x+3)(x^2+3x+9+x-9)=0
    ⇔(x+3)(x^2+2x)=0
    ⇔(x+3)x(x+2)=0
    ⇔[(x=-3),(x=0),(x=-2):}
    Vậy S={-3;0;-2}

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222-9+11+12:2*14+14 = ? ( )

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